Experiment No. 4
Linear and Non-Linear Systems
By:
Prof. Gabriel M. Rebeiz
The University of Michigan
EECS Dept.
Ann Arbor, Michigan
Any system (amplifier, filter, circuit, etc.) which has a transfer function:
Vo = A Vi where A ≡ H(ω)
is a linear system. The transfer function, A, can have a different value and phase for different
frequencies as we have measured in the lab (20 Hz - 20 kHz audio amplifier, low-Q and highQ filters, etc.). Linear systems have several basic properties which make them desirable for
use in electric circuits. Some of these properties are:
1. Linearity: If the input is a signal a1 and the output signal is b1(t), then if the input is na1(t),
the output will be exactly nb1(t) (n = constant). (At any frequency, Vo/Vi is a straight line!).
2. Superposition: If the input is a signal a1 and the output signal is b1, and if the input signal
is a2 and the output signal is b2, then if the input is a1 + a2, the output will be exactly b1 +
b2.
Talking in frequency domain, the input signal contains a frequency f, then the output will have
the same exact frequency f, and the only difference between the input and output is an
amplitude and phase change. If the input signal contains several frequencies (f1, f2, f3), then
the output signal will have exactly the same frequencies (f1, f2, f3) each changed differently in
its amplitude and phase, depending on the transfer function. However, no new frequencies are
generated in linear systems.
A non-linear transfer function is represented by:
Vo = A Vi + B Vi2 + C Vi3 + ...
where A, B, C are dependent on frequency
(A ≡ H(ω), B ≡ B(ω), C ≡C(ω) ...)
The above mentioned properties of linearity and superposition do not apply in non-linear
systems. In non-linear systems, if the input signal is a1(t), and the output signal is b1(t), then if
the input is na1(t), the output is not necessarily nb1(t)! Figure 1 shows an amplifier operating
in the linear and non-linear regime. In the non-linear region, the output is constant
independent of the input! As is evident, in a non-linear system, the output does not have the
same form as the input in time domain. This means that if the input signal contains a
frequency f, then the output signal will not only contain f but also 2f, 3f, ..., each at a different
amplitude depending on B(ω), C(ω), etc. The 2f, 3f, ... components are called harmonics of the
signal. They are generally non-desirable components in amplifiers, filters, etc. (systems which
should act linearly). Remember in EECS 210 how an amplifier generates a lot of harmonics
when driven into clipping (non-linear behavior)!
1
Real-Life Amplifiers:
In real life, linear amplifiers have some very small non-linear components (that is, the straight
line is not perfectly straight). In audio amplifiers, it is important to keep the non-linear
components (the harmonics) below -40 dB of the fundamental component. This translates to a
total harmonic distortion of:
THD ≤
Σ Pharmonics
≤
Psignal
0.0001
≤ 1% .
1
Vo
Linear
nb1
Non-Linear
b1
Slope
Vo/Vi = A
Vi
a1
Case 1
a1
Case 2
na1
na1
Since most amplifiers clip asymmetrically, they generate only Vi3, Vi5, etc. components. The
3
5
transfer function is therefore Vo = AVi + BVi + CVi + ... . If the Vi3 is the largest
component, then for a THD of 1%, B = 0.01 ≡ - 40 dB (for A = 1, or B/A = 0.01 for A ≠ 1).
Some people can hear this distortion level and hi-fi audio amplifiers are designed to give a
THD of 0.1% and even 0.01%. For a THD of 0.01%, B = 0.0001 ≡ - 80 dB (for A = 1, or B/A =
0.0001 for A ≠ 1), which is a -80 dB third harmonic component compared to the fundamental!
In communication systems, especially those radiating KWs of power such as radio stations, TV
stations, radars, etc., it is important to keep the system very linear and to generate very low (60 to -80 dB is very common) harmonic levels so as not to interfere with other stations at the
harmonic frequencies. This means if a TV station is radiating 50 KW at 200 MHz with a -60 dB
harmonic content, it will not radiate more than 50 mW(!) at 400 MHz, 600 MHz, etc. For low
power applications (0.2-1W) such as hand-held analog telephones at 50 MHz and digital
phones at 800-900 MHz, the harmonic content is about -30 dB.
Operation of Diodes in Small-Signal (Linear) Regime:
A diode is a very non-linear device. The 1-V curve is exponential and is given by
I = Is (e V n VT −1). However, as discussed in class, if a small-signal, Vs, is applied across the
diode around a DC bias condition of (ID, VD), then the diode equation can be written as:
2
2
3
⎛ V
⎞
1 ⎛⎜ Vs ⎞⎟
1 ⎛⎜ Vs ⎞⎟
s
Id = I D ⎜
+
+
+ ...⎟ .
2! ⎝ nVT ⎠
3! ⎝ nVT ⎠
⎝ nVT
⎠
The small signal (linear component) is
id = I D
Vs
, which is, of course, at the same
nVT
frequency as Vs. The other components, Vs2, Vs3, ... are non-linear components and
generate higher order harmonics. In class, we said that Vspk < 5 mV for “small-signal” or
“linear” operation. Let us now calculate the harmonic content for different values of Vspk.
Vs = A cos (ωt )
(V
spk
= A)
Id
⎛ A cos (ω t ) 1 A 2 cos2 (ω t ) 1 A3 cos3 (ωt )
⎞
Id = I D ⎜
+
+
+ ...⎟
2
2
3
3
2
n VT
6
n VT
⎠
⎝ nVT
Fundamental
Component:
Vs
–
negligible for A << nVT
3
Id(f) ≈
ID A
1 ID A
cos (ωt) +
cos (ωt )
nVT
8 n 3 VT 3
Second Harmonic (and DC!) Components:
I D A2
I d(2f) ≈
cos (2 ωt )
4n 2 VT 2
2
Id(DC) =
1 ID A
4 n 2 VT 2
Look: a DC component!
Third Harmonic Component:
Dividing, we have:
I(2 f )
I( f )
and
3
+
ID A 3
I d (3 f ) ≈
cos (3ω t )
24 n3 VT 3
1 A
≈
4 (nVT )
and
I(3 f )
I( f )
1 ⎛⎜ A ⎞⎟
≈
24 ⎝ nVT ⎠
2
nVT = 30 mV
Vspk = A I2f (dB) I3f (dB) THD
(mV) I
If
f
1
5
10
− 42
−28
−22
−87
< 1%
−59
−47
~ 4%
~ 8%
with Vspk being the voltage across the diode junction.
It is seen that the second and third harmonic levels are strongly dependent on the peak input
voltage (A). So, by limiting Vspk (= A) < 5 mV, we ensure that all harmonic are less than -28
dB and that the THD is less than 4%. This may simply not be enough in many applications,
and we may limit Vspk ≤ 1 mV if we want the harmonics to be less than -40 dB.
Non-Linear Systems Put To Good Use:
Do not think that non-linear systems are all bad! In many industrial problems, the process is
non-linear and it is best to use a non-linear system to control it. Also, most biological sensors
are non-linear (your eye, ear, pain sensors immediately go non-linear and saturate if too much
light, sound, or pain is applied). Finally, in communication systems, many non-linear devices
are expressly used to translate frequencies from 800-900 MHz to 10-20 MHz (and vice versa).
These components are called “mixers” and “multipliers” and are used in every communication
system today. You will study them in detail in EECS 411 and 522.
4
Experiment No. 4
Determining Diode Parameters
I
V nVT
Is e
V
I
Is e
1
+
IR s
nV T
I +
V –
Rs
V
with Rs ~ 0
–
I
1
with Rs
V
0
V
with Rs
V
Is e V nVT
I
V
n
I s nVT
I
log10 I2
Limiting due
to Rs
Slope
n
log10 Is
V
V1
V2
0
Is
VT ,
Rs ~ 0
log10 I1
I
VT
V3
nVT
nI
nI s
V4
Do a n I to log10 I change
V (linear)
Slope of the curve
nVT
V2 V1 2.3 nVT log 10 I2
Get n
log 10 I1
At large currents, voltage drop is series resistance limited.
⎛ V − IRs ⎞
I
⎜
= e nVT ⎟
⎟
Is ⎜⎝
⎠
and
ln
V − IRs
I
=
nVT
Is
⎛
I⎞
⇒ V = ⎜⎜ nVT ln ⎟⎟ + IRs
Is ⎠
⎝
V4 − V3 = nVT [ln (I4 ) − ln (I3 )] + (I4 − I3 ) Rs
if
5
I4 ≈ Is
then
V4 − V3
~ Rs
I4 − I3
⇒ The "flatter" the curve the
better the approximation.
Good at large currents.
Experiment No. 4
Amplitude Modulation, Envelope and Square-Law
Diode Detectors
1.0 Amplitude Modulation:
An amplitude modulated (AM) wave is given by:
V (t ) = {A + a(t )}sin (ω c t )
where
signal)
a(t) ≡ audio signal (commonly called baseband signal, or modulation
ωc ≡ carrier frequency
A ≡ Amplitude of carrier frequency for no modulation signal
For the AM radio broadcast, the carrier frequency is between 550 KHz and 1550 KHz,
and the bandwidth of the audio signal is 5 KHz. The audio signal is composed of a
range of frequencies from 20 Hz to 20 KHz and each frequency is associated with a
certain amplitude, resulting in the audio spectrum for a specific instrument or human
voice. Most of the power of the human voice is below 5 KHz, so AM is excellent for
talk-shows and not for music.
Let us assume that the audio signal (or baseband signal, or modulation signal) is
composed of a single frequency, ω1 (for example, ω1 = 1 KHz). The AM signal is then:
V (t ) = {A+ a1 sin (ω1 t )}sin (ω c t )
= A sin(ω c t ) + a1 sin (ω1t )sin (ω c t )
= A sin (ω c )t −
a1
a
cos (ω c + ω1 )t + 1 cos (ω c − ω1 )t
2
2
We see that the AM signal is composed of three frequencies: the carrier frequency
(ωc) with amplitude A, the upper sideband frequency (ωc+ω1) and the lower sideband
frequency (ωc-ω1), each with amplitude a1/2. An AM signal with a1=0.9A is shown in
Fig. 1. The peak of the signal is 1.9A and the minimum is 0.1A. The frequency
spectrum of the signal is also shown for fc= 1 MHz and ω1= 5 KHz. Notice that the
bandwidth of an AM signal (centered around the carrier) is twice the bandwidth of the
baseband signal and in this case is 10 KHz.
An often used term is the modulation index, m, defined by m = a1/A. Using this
definition, the AM signal can be written as:
V (t ) = A(1+ m sin (ω 1t ))sin (ω ct )
m
m
= A ⎡ sin (ω c t ) − cos (ω c + ω 1 )t + cos (ω c − ω1 )t ⎤
⎣
⎦
2
2
The total power contained in an AM signal is:
⎛ A2 ⎞
⎜
⎟ (1+ m2 2)
Ps =
⎝ 2 Re {Z}⎠
6
where Z is the impedance seen by the AM signal, and Pc is the power in the carrier
frequency (Pc=A2/2Re{Z}). The sidebands contains only m2/2 of the power relative to
the carrier signal power. For m=1 (100% modulation), Ps=Pc(1+1/2) and the
sidebands contain half of the carrier power, and one third of the total AM signal power.
Amplitude modulation is therefore not an efficient method for transmitting information,
but it is so easy to generate and to detect that it is the predominant modulation
technique in low-cost low-bandwidth systems (in other words, no one uses AM
modulation for satellite and wireless communications). Most AM broadcast stations
maintain a modulation index, m, between 85 and 95% as a compromise between
spectral efficiency (power in baseband vs. power in carrier) and the risk of drifting into
overmodulation and therefore severe distortion (m > 100%).
Carrier
Sidebands
dB
c-
1
c
c+
1
f
Figure 1: An amplitude modulated (AM) signal in time and frequency domain.
AM modulation is not limited to radio broadcasts and to audio signals. Actually, when
you use a TV remote control, or an infrared transmitter on a calculator, or an infrared
data link between your computer and a hard disk, you are actually AM modulating the
infrared diode with a digital (on, off) modulation. In this case, the carrier is at an
extremely high frequency (~300 THz for a red diode) and the baseband signal is a
digital signal with a frequency of 100’s of Hertz for the case of the TV remote control to
several MHz for the case of a hard disk data link. We will build an AM infrared link in
Experiment #5.
2.0 A Square-Law Detector:
If an AM signal (or any type of signal) passes by a linear system with
vo=Avi (A=constant), then the output is a replica of the input (in time and frequency
domain). The output differs from the input only by its amplitude and phase but does
not contain any new frequency components. The amplifiers are called linear amplifiers
(such as audio amplifiers, communication amplifiers, etc. ...) and generate very little
total harmonic distortion (i.e., no harmonics, no new frequency components). This is
the essential characteristic of linear systems which I tried hard to get across in EECS
210. Therefore, linear systems cannot detect modulation of any type!
If one needs to detect a signal, a frequency translation must occur. Basically, the
information in the sidebands at (ωc+ω1) and at (ωc-ω1) must be translated to ω1! For
example, the information in the 1005 KHz and 995 KHz sidebands of Figure 1 must be
translated into 5 KHz so as to be heard as an audio signal. In order to do this, the
detector must have a non-linear transfer function. The simplest non-linear transfer
function is the square-law detector. Let us assume that a detector exhibits a response
given by:
Vo = k Vi2 k = constant, and Vi (t ) = A (1+ m sin (ω 1t )sin (ω ct )) (the AM signal)
7
The output voltage is:
kA2 kA2 m2
Vo (t ) =
+
2
4
2
+ kA m sin (ω 1t )
kA2 m2
+
sin (2 ω1 t )
4
2
kA2
+
cos(2ω c t )[1+ m sin (ω1 t )]
2
DC Component
Audio Component
Distorted Audio Component
Components
at
2ωc
with
respective
sidebands
Notice that we have frequency translation! An input frequency spectrum of ωc, ωc+ω1
and ωc-ω1 (the AM signal) results in an output frequency spectrum containing a DC
component, a baseband (or modulation or audio) component at at ω1, a component at
2ω1 (called distortion), and frequency components 2ωc with respective sidebands. If
you remember EECS 210, we called these components harmonics and
intermodulation products and they occurred whenever an amplifier was driven into the
non-linear region (clipping). Well, here we are using the intermodulation components
of a square-law detector to our advantage so as to translate the sidebands at ωc + ω1
to an audio frequency of ω1.
3.0 AM Signal Detection (or Demodulation):
The question is then: which component do we use to get a good square-law
response? An excellent and very inexpensive electronic component is the pn junction
diode (or the Schottky diode for high frequency applications, f > 100 MHz). The diode
IV relationship is given by:
I = Is (e
αV
−1), α = 1nV
T
with Is= Diode Saturation Current, VT= 26 mV and
n=1.2–1.8 for silicon diodes and germanium diodes.
If we expand the exponential equation into its Taylor series, we get:
⎛
α 2 V2
⎞
I = Is ⎜ 1+ α V +
+ ... −1
⎝
⎠
2
= α Is V +
α 2 I s V2
2
+ V
–
I
+ ...
and we can put it in the form:
I = k1V + k2 V 2 + ... with
Now,
let
us
k1 = α Is and
assume
k2 =
that
α 2 Is
an
2
AM
voltage,
Vi (t ) = VAM (t ) = A (1 + m sin(ω1 t ))sin (ω c t ), is impressed across the diode, it does
not take a genius to recognize that the V2 component in the diode current equation
will demodulate the AM signal! The resulting current in the diode is:
8
2
I = k1 VAM (t ) + k2 VAM
(t ) +...
Expanding the sine/cosine terms and grouping them together, we get the diode
current I to be:
2
k2 A 2
2 m
I=
+ k2A
2
4
2
+ k2 A msin (ω 1t )
A2 m2
+ k2
sin (2ω 1t )
4
+ k1 A (1+ m sin (ω 1t ))sin (ω ct )
DC Component
Audio Component
Distorted Audio Component
Components
at
ωc
and
Components
at
2ωc
and
sidebands
2
k2 A2
+
1+ m sin (ω1 t )) cos (2ω c t )
(
2
sidebands
+
VAM
–
I
~
Notice that the diode current has a DC component and therefore the diode self-biases
at
(k
2
A 2)(1 + m 2) (Amperes). This is the result of the non-linear (demodulation)
2
2
process.
The diode current is then passed by a parallel RC network (a low-pass filter). This
network presents an resistance R at low frequencies (ω<<1/RC) and a short circuit a
high frequencies (ω >>1/RC).
+
I
VAM
C
–
+
R
Vo
–
Figure 2: AM demodulation circuit (envelope detector).
9
The resulting output voltage is therefore:
=R
Vo
k2 A
2
2
2
k2 A m
4
2
R||C
~R
2
k 2 A m sin
1
R||C
t
1
~R
2
2
A m
k2
sin 2 1 t
4
k 1 A 1 m sin
c
R||C
t sin
c
t
2
1
R||C
c
~ 0!
(negligible component in Vo)
~ 0!
. . . . . . . . . . . cos 2
c
t
R||C
2
(negligible component in Vo)
c
Notice that the output voltage has a DC component, a baseband components at ω1
and distortion components 2ω1. The DC component can be removed with a simple
series capacitor and voila! we have the baseband (audio) signal. The AM
demodulation circuit is commonly called an “envelope” detector since it follows the
envelope of the AM signal.
2
The distortion signal at 2ω1 is dependent on m 4 , and the audio signal is dependent
on m. For high fidelity AM, m should be kept around 0.5 which results in a distortion
component of around -20 dB with respect to the audio component. In this case, the
sidebands contain only 12% of the power of the carrier which is not an efficient use of
the radio power spectrum. Therefore, it is common to use m ~
– 0.9 for AM broadcast
of human voice which results in a distortion component around -13 dB relative to the
signal.
For audio links using AM modulated infrared systems, spectral power is of no concern
since the distances are generally short (for ex: TV room, conference room). In this
case, m is kept around 0.2-0.3 for low distortion response in analog systems.
10
Another Point of View:
A simple way to explain the AM detector circuit which is common in radio amateur
circles is presented below: Consider the circuit of Figure 2 with an impressed AM
voltage of Vi(t) = A(1+msin(ω1t))sin(ωct). The capacitor is chosen to be a short-circuit
at the carrier frequency and therefore vi(t) is imposed across the diode. The diode
rectifies the signal, much like a half-wave bridge rectifier, and the capacitor charges to
the peak of the signal. The resistor is chosen so as to discharge the capacitor slowly,
and therefore the output waveform follows the envelope of the AM modulated signal.
The output waveform has a DC component, and a series capacitor is used before the
baseband (audio) amplifier to block this DC voltage. This is illustrated in Figure 3 (from
the Radio Amateur’s Handbook).
AM detectors are generally driven from low impedance sources since the non-linear
“action” is due to the diode current. A low source impedance ensures a large current in
the diode for a given source voltage. This is the reason why the AM detector is placed
after an op-amp (low output impedance), after the emitter of a transistor (very low
impedance) or after a transformer which ensures a low impedance drive.
Problems with AM detection:
The simplicity of AM detection is also its downfall. Anything which exhibits a non-linear
response can demodulate an AM signal! In an AM receiver, one must be careful not to
drive any of the pre-amplifiers into saturation (clipping), otherwise, “free and
unwanted” AM demodulation occurs. The non-linear response is not limited to
electronic circuits. Speakers can demodulate AM signals and I have heard many a
radio amateur voice popping out on my speakers (speaker coils are non-linear)! One
interesting true story is about this person who had two metals in his tooth (gold and
silver). The junction between these metals with the conducting alkaline fluid (saliva) in
his mouth created a non-linear diode and therefore demodulated AM stations. Since
this happened in his mouth, the demodulation process did not need to be efficient, and
he could hear music and talk shows whenever he drove by an AM station!
11
Fig. 3: Radio’s simplest demodulator, the diode rectifier (A), demodulates an
AM signal by multiplying its carrier and sidebands to produce frequency sums
and differences, two of which sum into a replica of the original modulation (B).
Modern receivers often use an emitter follower to provide low-impedance drive
for their diode detectors (C).
12
Experiment No. 4
Diode Small-Signal Measurements, Amplitude
Modulation,
Envelope and Square-Law Diode Detectors
Goal: To measure the small-signal characteristics of a diode. To learn about AM signals
and to build an AM demodulator using a square-law diode detector.
Read this experiment and answer the pre-lab questions.
Equipment: • Agilent E3631A Triple output DC power supply
• Agilent 33120A Function Generator
(Replacement model: Agilent 33220A Function / Arbitrary Waveform
Generator)
• Agilent 34401A Multimeter
• Agilent 54645A Oscilloscope
(Replacement model: Agilent DSO5012A 5000 Series Oscilloscope)
1.0 Silicon and Germanium Diodes:
As discussed in class, germanium (Ge) diodes have a much lower turn-on voltage
than silicon diodes and therefore are very useful for small-signal AM detection.
1. Set the Agilent E3631A to give a maximum current of 100 mA for the +6 V supply.
2. Connect the 1N34A Ge diode as shown below to measure the I-V curve.
Measure the diode current for V= 0.0 - 0.32 V in 0.04 V steps and at 0.4 – 0.8
in 0.1 V steps.
3. Connect now the 1N4148 silicon diode.
Measure the diode current for V = 0.3–1.4 V in 0.1 V steps.
2.0 Diodes and Small-Signal Performance:
Connect the Ge 1N34A diode to a +5V supply with a 4.7 KΩ resistor and to the signal
generator using a 4.7 µF DC-block capacitor and a 560 Ω series resistance.
13
1. a.
Draw the diode circuit, together with the equivalent DC, AC/circuits of the
1N34A diode in your notebook.
b.
Measure the DC voltage across the diode (VoDC) and calculate the dc
diode current (ID).
c.
Calculate the diode ac resistance (rd) for n = 1.3.
2. Set the Agilent 33120A function generator to deliver a 4 KHz sinewave with
Vppk = 400 mV. Connect it to the 1N34A diode circuit using a 560 Ω series
resistor and a 4.7 µF series capacitor. The 4.7 µF capacitor is a simple DC-block.
The 610 Ω (560 + 50 Ω) forms a voltage divider at the input with the small-signal
resistance of the diode.
3. a. Connect Vo to the scope and plot the ac voltage across the diode (label
Voppk and Vo(av)). This is the ac voltage across the total diode resistance (rd
+ Rs).
b. Plot the output voltage (Vo) in the frequency domain and write the level of the
fundamental and second/third harmonic components? Is the diode operating in
the linear region? (In this case, linear region means that the harmonics are -20 dB
or less compared to the fundamental.)
c. Set now Vs = 1Vppk and plot again the voltage across the diode (time and
frequency domain). Is the diode still operating in the linear region?
14
3.0 AM Modulation
1. Connect the Agilent 33120A signal generator to the Agilent scope. Connect the
sync. signal of the signal generator to the external trigger input. Set the scope
trigger to External.
AM
Shift
.
2. To enter into AM mode, press
AM
Shift
(To leave the AM mode, press again
at any time).
3. To set the carrier frequency:
This is simple. Just set the frequency (sinewave) at 100 KHz with Vppk = 2V.
(IMPORTANT NOTE: The Agilent 33120A signal generator delivers half of Vppk
shown on the display when it is in the AM mode. Therefore, the actual carrier
Vppk measured on the scope is 1 V.)
4. To set the Modulation Frequency:
Freq
Shift
Freq
, and enter the carrier frequency using the knob or the
Press
COMMAND. Enter 5 KHz.
5. To set the Modulation Level:
Enter
Number
Level
Shift
Ampl
and using the knob, set the modulation level (from 0% to 120%
Press
in 10% steps). Enter 40%.
6. a.
Choose a long enough timebase to see the AM waveform on the screen
through several cycles. Draw the envelope on your lab notebook. Label
the peak and minimum amplitudes of the envelope.
b.
Go into the FFT mode and choose a center frequency of 122 KHz, and a
frequency span of 244 KHz. Measure the frequency of the carrier and the
sidebands, and their amplitude levels in dB.
7. Set the modulation level (m) to 80% and 120%. In each case, plot approx. (do not
take data) the envelope in time domain, and the frequency spectrum in FFT
domain. What do you notice when m > 100%?
(The reason I chose 122 KHz for the carrier frequency and 5 KHz for the
modulation frequency is that the spectrum analyzer (FFT mode of the scope) does
not have a 1% resolution. This means that it would be very hard to see the
sidebands with a 500 KHz or a 1 MHz carrier frequency).
4.0 AM Demodulation (Envelope Detector):
1. Set the AM signal to have the following specs:
a. Carrier Frequency: 1 MHz and 1 Vppk on the scope. Remember: This is AM
mode and you should enter 2 Vppk for the carrier.
b. Modulation Frequency/Level: 1 KHz and m = 40%.
c. Look at the AM signal on the scope and make sure that everything is OK.
Measure the peak and minimum amplitudes of the waveform envelope.
They should be around 0.7 V and 0.3 V, respectively.
2. Connect the AM signal to the series diode as shown below.
a.
Draw the series rectifier circuit in your notebook.
b.
Plot Vs and Vo on your notebook (you will see rectification). Determine
the diode voltage drop.
15
3. Connect the AM signal to the envelope detector or AM demodulator (add the 4.7
nF capacitor) as shown below:
a.
b.
Draw the AM demodulation circuit in your notebook.
Plot the demodulation signal Vo in time domain and determine its
frequency (1 KHz), DC level and Vppk.
c.
From the DC voltage level, calculate the DC rectified current in the diode
knowing that Vodc = IdcR. If you wish, you may remove the DC level from
the scope by choosing an AC input option in channel 1.
d.
Plot the spectrum of Vo (FFT) with a frequency span of 9.6 KHz and
determine the value of the fundamental (1 KHz) and distortion component
(2 KHz, 3 KHz, ...). The distortion components should be very low and
may be not measurable (make sure that you are not clipping the signal on
the scope).
e.
Return to time domain: Using the knob, vary quickly the modulation
frequency (fm) from 300 Hz to 30 KHz and determine the 3-dB bandwidth
of the demodulator (that is, when Voppk drops by 0.707). Do not measure
the frequency response! Return to fm= 1 KHz.
f.
Using the knob, change the carrier frequency from 450 KHz to 1650 KHz
(this is the AM band on your radio) and look at the demodulated 1 KHz
signal. What do you notice ? Is the demodulated signal constant vs.
carrier frequency? Return to fc = 1 MHz.
4.
Repeat 3(b), and 3(d) for a modulation level of m = 80% and m = 120% (if you
have time). (Measure the third harmonic distortion component if possible).
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Experiment No. 4
Amplitude Modulation, Envelope and Square-Law
Diode Detectors
Pre-Lab Assignment
1. A silicon diode with n = 1.2, Rs=10 Ω is biased from a 9 V supply using a 4.7 KΩ bias
resistor. The DC voltage across the diode is 0.7 V.
As mentioned in class, the battery looks like a large capacitor and therefore acts as a
short circuit in the ac model (+9 V battery ≡ ground).
a. Calculate the diode ac resistance (rd=nVT/ID), where ID ≡ diode DC current. Calculate
the total diode resistance (Rs + rd). Note that the ac circuit is superimposed on the DC
circuit and Vo=0.7 V + vo, and Id=Id+id, where vo, id are the ac voltages and currents
of the diode.
b. The diode is fed by a 200 Ω ac source with Vpk=100 mV. Calculate the ac voltage
across the diode (across Rs + rd), the ac current in the diode (id) and the voltage
across the diode junction (across rd).
As mentioned in class, the peak ac voltage across the diode junction should be less
than 10mV so that the diode operates in the “linear” region. Is the diode above
operating in the linear region?
c.
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A 20 KΩ bias resistor is now used instead of the 4.7 KΩ resistor. Repeat a-b. Will the
diode be still operating in the linear region? Explain.
2. For an AM signal, calculate the power in each sideband frequency relative to the carrier
for m = 40%, 80% and 120%.
3. An AM signal (fc = 800 KHz, f1= 3 KHz, m =80%) with A=1 Vpk is fed to a square-law
diode detector (envelope detector) with Is= 40 nA , VT = 26 mV and n = 1.5.
a. Calculate the peak and minimum amplitudes of the AM waveform (envelope).
b. Calculate all the frequencies which are available in the current spectrum.
c.
Calculate the values of the output voltage frequency components (DC, baseband and
distortion) for a load R = 10 KΩ, C = 4.7 nF.
4. In the AM demodulator circuit, what will happen if the diode orientation is reversed?
Consider both the DC and ac (at f1) components of the output waveform.
Optional:
You are welcome to review the Agilent AM Fundamental Computer-Based Training package.
Talk to your TA about it.
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Experiment No. 4
Amplitude Modulation, Envelope and Square-Law
Diode Detectors
Lab-Report Assignment
1. Plot the 1N4148 and 1N34A diode I,V curve on the same (log I, V) graph. Calculate n and
Rs for both diodes using the handout given in class for the silicon and Ge diodes.
2. a. Summarize in a table your results for the Ge 1N34A diode small-signal experiment
(section 20) such as n, Rs, rd, Vo(DC), Voppk, Vfo and V2fo, V3fo, etc. for Vs=400
mVppk and Vs=1 Vppk.
b. What is the ac current and ac voltage in the diode across the junction (across rd) in
each case? Confirm that the diode operates in the linear region (nearly) for Vs=400
mVppk and in the non-linear region (definitely) for Vs=1 Vppk.
c. Calculate the high-pass corner frequency of the 4.7 µF capacitor. (Use the ac circuit
and determine the equivalent resistance seen by the 4.7 µF capacitor).
3. a. Summarize in a table your results on the AM envelope detector for m = 40%, m = 80%
and m =120% (VODC, Voppk, rectified current IDC, fundamental voltage and
distortion component, corner frequency for m = 40%). State the value of A, and your
carrier and modulation frequencies.
b. Calculate the expected corner frequency of the AM detector for m = 40% knowing IDC
and therefore the ac resistance of the diode. Compare to your measurements.
Comment. (To calculate the corner-frequency, use the ac circuit and determine the
equivalent resistance seen by the 4.7 nF capacitor. You can find the equivalent
resistance of the diode by knowing the DC output voltage and hence the DC current in
the diode).
These experiments have been submitted by third parties and Agilent has not tested any of the experiments. You will undertake any of
the experiments solely at your own risk. Agilent is providing these experiments solely as an informational facility and without review.
AGILENT MAKES NO WARRANTY OF ANY KIND WITH REGARD TO ANY EXPERIMENT. AGILENT SHALL NOT BE LIABLE FOR
ANY DIRECT, INDIRECT, GENERAL, INCIDENTAL, SPECIAL OR CONSEQUENTIAL DAMAGES IN CONNECTION WITH THE USE
OF ANY OF THE EXPERIMENTS.
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